Theorem 1

Let $A$ be a commutative ring, $M$ a free $A$-module and $(e_\lambda)_{\lambda \in L}$ a basis of $M$. Then the elements $e_s = e_{\lambda_1} \otimes e_{\lambda_2} \otimes \cdots \otimes e_{\lambda_n}$, where $s = (\lambda_1, \ldots, \lambda_n)$ runs through the set of all finite sequences of elements of $L$ and $e_\varnothing$ is used to denote the unit element of $T(M)$, form a basis of the $A$-module $T(M)$.

The elements of this basis are obviously homogeneous and the multiplication table is given by

where $st$ denotes the sequence of elements of $L$ obtained by juxtaposing the sequences $s$ and $t$ (I, § 7, no. 2).

It is seen that the basis $(e_s)$ of $T(M)$, with the multiplicative law (10), is canonically isomorphic to the free monoid of the set $L$ (I, § 7, no. 2), the isomorphism being obtained by mapping each word $s$ of this monoid to the element $e_s$. It follows (§ 2, no. 7) that the tensor algebra $T(M)$ of a free module $M$, with a basis of indexing set $L$, is canonically isomorphic to the free associative algebra of $L$ over $A$. In particular (§ 2, no. 7, Proposition 7), for every mapping $f : L \to E$ from $L$ to an $A$-algebra $E$, there exists one and only one $A$-algebra homomorphism $\bar{f} : T(M) \to E$ such that $\bar{f}(e_\lambda) = f(\lambda)$.